Question

sum of 3 consecutive terms of an AP is 42 and product of 1st and last term is 180.find the terms.

Answer 1

Let the three consecutive terms be (a-d) (a) & (a+d) resp.

(a-d) +(a) +(a+d) =42
3a=42
a=14
(a-d)(a+d)=180
a²-d²=180
(14)²-d²=180
d²=196-180
d²=16
d=±4

Thus If d=+4,then the three terms are 10,14 and 18 resp and if d=-4 the three terms are 18, 14 and 10 resp.

Answer 2

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